Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

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Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:

import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;

/**
 * Created by imahmoud on 5/13/17.
 */
public class C {

    public void solve(){
        Scanner in =  new Scanner(System.in);
        int N1 = in.nextInt();
        int N2 = in.nextInt();


        int[][] grid;
        int W,H;
        boolean rotated =false;
        if(N1 < N2){
            W = N1;
            H = N2;
            grid = new int[W][H];
            for (int i = 0; i < N1; i++) {
                for (int j = 0; j < N2; j++) {
                    grid[i][j] = in.nextInt();
                }
            }
        }else{
            rotated = true;
            W = N2;
            H = N1;
            grid = new int[W][H];
            for (int i = 0; i < N1; i++) {
                for (int j = 0; j < N2; j++) {
                    grid[j][i] = in.nextInt();
                }
            }
        }


        int min = Integer.MAX_VALUE;
        for (int j = 0; j < H; j++) {
           min = Math.min(min,grid[0][j]);
        }
        List<String> res = new ArrayList<>();
        for (int i = 0; i < H; i++) {
            int diff = grid[0][i] - min;
            for (int j = 0; j < diff; j++) {
                res.add((rotated ? "row " : "col ")+ (i+1));
                for (int f = 0; f < W; f++) {
                    grid[f][i]--;
                }
            }
        }

        for (int i = 0; i < W; i++) {
            int diff = grid[i][0];
            for (int j = 1; j < H; j++) {
                if(grid[i][j] != diff){
                    System.out.println(-1);
                    return;
                }
            }
            for (int j = 0; j < diff; j++) {
                res.add((rotated ? "col ": "row ")+ (i+1));
            }
        }

        System.out.println(res.size());
        for (int i = 0; i < res.size(); i++) {
            System.out.println(res.get(i));
        }

    }
    public static void main(String[] args) {
        new C().solve();
    }
}